Fluid Dynamics PYQs: MHT CET, JEE Main, NEET

Q1MHT CET 2026 · 20 APR · SHIFT 1Easy

Types of Flow, Reynolds Number & Continuity · Types of flow & Reynolds number

In case of liquid, if Reynold's number ' $R_n$ ' is $1450$ , then the flow of liquid will

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Answer: (c)

Reynolds number ranges: $R_n < 1000$ streamline, $1000 < R_n < 2000$ unsteady (changing), $R_n > 2000$ turbulent.
$R_n = 1450$ lies between $1000$ and $2000$ , so the flow is in transition from streamline to turbulent.
So the answer is (c).

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Q2MHT CET 2026 · 18 APR · SHIFT 2Easy

Types of Flow, Reynolds Number & Continuity · Types of flow & Reynolds number

In case of fluid flowing on horizontal surface the flow of fluid becomes unsteady when Reynolds number (Rn) is

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Answer: (d)

Reynolds number ranges: $R_n < 1000$ streamline, $1000$ to $2000$ transition, $R_n > 2000$ turbulent.
Fully irregular (unsteady, turbulent) flow sets in when $R_n > 2000$ .
So the answer is (d) greater than 2000.

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Q3MHT CET 2026 · 19 APR · SHIFT 1Easy

Types of Flow, Reynolds Number & Continuity · Equation of continuity

An in-compressible fluid flows steadily through a horizontal cylindrical pipe. The pipe has radius ' $3R$ ' at point A and ' $1.5R$ ' at point B, further along the flow direction at the same level. If the velocity at point A is ' $v$ ', then that at point B is

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Answer: (d)

Continuity: $A_A v_A = A_B v_B \Rightarrow r_A^2 v = r_B^2 v_B$
$v_B = v\left(\dfrac{3R}{1.5R}\right)^2 = 4v$
So the answer is (d) $4v$ .

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Q4MHT CET 2026 · 16 APR · SHIFT 1Easy

Types of Flow, Reynolds Number & Continuity · Equation of continuity

Water flows through a horizontal pipe at a speed 'V'. Internal diameter of the pipe is 'd'. If the water is emerging at a speed ' $V_1$ ' then the diameter of the nozzle is

______

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Answer: (a)

Continuity: $\dfrac{\pi d^2}{4}V = \dfrac{\pi D^2}{4}V_1$ , where $D$ is the nozzle diameter.
$D^2 = d^2\dfrac{V}{V_1} \Rightarrow D = d\sqrt{\dfrac{V}{V_1}}$
So the answer is (a) $d\sqrt{V/V_1}$ .

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Q5JEE Main 2026 · 4 APR · SHIFT 2Easy

Types of Flow, Reynolds Number & Continuity · Equation of continuity

A water spray gun is attached to a hose of cross sectional area $30\text{ cm}^2$ . The gun comprises of 10 perforations each of cross sectional area of $15\text{ mm}^2$ . If the water flows in the hose with the speed of $50\text{ cm/s}$ , calculate the speed at which the water flows out from each perforation. (Neglect any edge effects)

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Answer: (b)

Continuity: $A_1 v_1 = n a v$ (all perforations together).
$A_1 v_1 = 30\text{ cm}^2 \times 50\text{ cm/s} = 1500\text{ cm}^3/\text{s} = 1.5\times10^{-3}\text{ m}^3/\text{s}$
$na = 10 \times 15\text{ mm}^2 = 150\text{ mm}^2 = 1.5\times10^{-4}\text{ m}^2$
$v = \dfrac{1.5\times10^{-3}}{1.5\times10^{-4}} = 10\text{ m/s}$
So the answer is (b) $10\text{ m/s}$ .

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Q6MHT CET 2025 · 21 APR · SHIFT 2Easy

Types of Flow, Reynolds Number & Continuity · Equation of continuity

Water flows through a horizontal pipe of varying cross-section at the rate of $\pi \times 10^{-1}\text{ m}^3/\text{s}$ . The velocity of water at a point where the radius of the pipe is 10 cm is $(\pi = 3.14)$

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Answer: (c)

Rate of flow $Q = Av = \pi r^2 v$ .
$v = \dfrac{\pi \times 10^{-1}}{\pi (0.1)^2} = \dfrac{10^{-1}}{10^{-2}} = 10\text{ m/s}$
So the answer is (c) $10\text{ m/s}$ .

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Q7MHT CET 2025 · 21 APR · SHIFT 1Easy

Types of Flow, Reynolds Number & Continuity · Equation of continuity

An incompressible fluid flows steadily through a cylindrical pipe having radius R at point A and $\left(\frac{R}{3}\right)$ at point B further along the direction of flow of liquid. If the velocity at point A is ' V ' then that at point B is

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Answer: (d)

Continuity: $\pi R^2 V = \pi (R/3)^2 v_B$
$v_B = 9V$
So the answer is (d) $9V$ .

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Q8MHT CET 2023 · 12 MAY · SHIFT 2Easy

Types of Flow, Reynolds Number & Continuity · Types of flow & Reynolds number

Select the WRONG statement from the following. In a streamline flow

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Answer: (a)

In streamline flow, the velocity at a given point stays constant with time, so (a) is wrong.
(b), (c) and (d) are true: speed below critical velocity, parallel layers, no random motion.
So the wrong statement is (a).

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Q9MHT CET 2023 · 11 MAY · SHIFT 2Easy

Types of Flow, Reynolds Number & Continuity · Types of flow & Reynolds number

A fluid of density ' $\rho$ ' is flowing through a uniform tube of diameter ' $d$ '. The coefficient of viscosity of the fluid is ' $\eta$ ', then critical velocity of the fluid is

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Answer: (b)

$R_n = \dfrac{\rho v_c d}{\eta} \Rightarrow v_c = \dfrac{R_n \eta}{\rho d}$
$v_c \propto \eta$ , and $v_c \propto 1/\rho$ , $v_c \propto 1/d$ .
So the answer is (b) directly proportional to $\eta$ .

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Q10MHT CET 2023 · 9 MAY · SHIFT 2Easy

Types of Flow, Reynolds Number & Continuity · Types of flow & Reynolds number

A fluid of density ' $\rho$ ' and viscosity ' $\eta$ ' is flowing through a pipe of diameter ' $d$ ', with a velocity ' $v$ '. Reynold number is

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Answer: (b)

Reynolds number = inertial force / viscous force $= \dfrac{\rho v d}{\eta}$ .
So the answer is (b) $\dfrac{d\rho v}{\eta}$ .

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Q11MHT CET 2023 · 10 MAY · SHIFT 1Easy

Types of Flow, Reynolds Number & Continuity · Equation of continuity

Water flows through a horizontal pipe at a speed ' $V$ '. Internal diameter of the pipe is ' $d$ '. If the water is coming out at a speed ' $V_1$ ' then the diameter of the nozzle is

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Answer: (b)

Continuity: $d^2 V = D^2 V_1$ , where $D$ is the nozzle diameter.
$D = d\sqrt{\dfrac{V}{V_1}}$
So the answer is (b).

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Q12JEE Main 2022 · 26 JUN · SHIFT 2Easy

Types of Flow, Reynolds Number & Continuity · Types of flow & Reynolds number

If $\rho$ is the density and $\eta$ is coefficient of viscosity of fluid which flows with a speed v in the pipe of diameter d, the correct formula for Reynolds number Re is:

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Answer: (c)

Reynolds number $R_e = \dfrac{\rho v d}{\eta}$ (dimensionless).
So the answer is (c).

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Q13JEE Main 2022 · 28 JUL · SHIFT 2Medium

Types of Flow, Reynolds Number & Continuity · Equation of continuity

A pressure-pump has a horizontal tube of cross sectional area $10\text{ cm}^2$ for the outflow of water at a speed of $20\text{ m/s}$ . The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is:

[given: density of water $= 1000\text{ kg/m}^3$ ]

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Answer: (d)

Force = rate of change of momentum $= \dfrac{dm}{dt}v = (\rho A v)v = \rho A v^2$ (water stopped by the wall).
$F = 1000 \times 10\times10^{-4} \times (20)^2$
$F = 400\text{ N}$
So the answer is (d) $400\text{ N}$ .

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Q14JEE Main 2022 · 29 JUN · SHIFT 1Medium

Types of Flow, Reynolds Number & Continuity · Equation of continuity

A block of metal weighing 2 kg is resting on a frictionless plane (as shown in figure). It is struck by a jet releasing water at a rate of 1 kgs ${}^{-1}$ and at a speed of 10 ms ${}^{-1}$ . Then, the initial acceleration of the block, in ms ${}^{-2}$ , will be:

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Answer: (c)

The jet loses its horizontal momentum on hitting the block: $F = \dfrac{dm}{dt}v$ .
$F = 1\text{ kg/s} \times 10\text{ m/s} = 10\text{ N}$
$a = \dfrac{F}{m} = \dfrac{10}{2} = 5\text{ m/s}^2$
So the answer is (c) 5.

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Q15MHT CET 2021 · 24 SEP · SHIFT 1Medium

Types of Flow, Reynolds Number & Continuity · Types of flow & Reynolds number

A glass tube of uniform cross-section is connected to a tap with a rubber tube. The tap is opened slowly. Initially the flow of water in the tube is streamline. The speed of flow of water to convert it into a turbulent flow is [radius of tube $= 1\text{ cm}, \eta = 1 \times 10^{-3}\,\frac{\text{Ns}}{\text{m}^2}, R_n = 2500$ and density of water $= 10^{3}\text{ kg/m}^3$ ]

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Answer: (b)

$R_n = \dfrac{\rho v D}{\eta} \Rightarrow v = \dfrac{R_n \eta}{\rho D}$ , with $D = 2r = 0.02\text{ m}$ .
$v = \dfrac{2500 \times 10^{-3}}{10^3 \times 0.02} = \dfrac{2.5}{20}$
$v = 0.125\text{ m/s}$
So the answer is (b) $0.125\text{ m/s}$ .

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Q16JEE Main 2021 · 16 MAR · SHIFT 2Medium

Types of Flow, Reynolds Number & Continuity · Types of flow & Reynolds number

What will be the nature of flow of water from a circular tap, when its flow rate increased from 0.18 L/min to 0.48 L/min? The radius of the tap and viscosity of water are 0.5 cm and 10 $-$ 3 Pa s, respectively. (Density of water: 103 kg/m3)

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Answer: (a)

$R_e = \dfrac{\rho v D}{\eta}$ with $v = \dfrac{Q}{\pi r^2}$ gives $R_e = \dfrac{2\rho Q}{\pi \eta r}$ .
$Q_1 = 0.18\text{ L/min} = 3\times10^{-6}\text{ m}^3/\text{s}$ : $R_e = \dfrac{2\times10^3\times3\times10^{-6}}{\pi\times10^{-3}\times0.005} \approx 382$
$Q_2 = 0.48\text{ L/min} = 8\times10^{-6}\text{ m}^3/\text{s}$ : $R_e \approx 1019$
$R_e$ goes from below $1000$ (steady) to above $1000$ (unsteady).
So the answer is (a) steady flow to unsteady flow.

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Q17JEE Main 2020 · 7 JAN · SHIFT 2Easy

Types of Flow, Reynolds Number & Continuity · Equation of continuity

An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is:

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Answer: (b)

Continuity: $v \propto \dfrac{1}{A} \propto \dfrac{1}{d^2}$ .
Minimum velocity is at maximum diameter: $\dfrac{v_{min}}{v_{max}} = \left(\dfrac{d_{min}}{d_{max}}\right)^2 = \left(\dfrac{4.8}{6.4}\right)^2$
$= \left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16}$
So the answer is (b) $\dfrac{9}{16}$ .

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Q18JEE Main 2019 · 8 APR · SHIFT 1Medium

Types of Flow, Reynolds Number & Continuity · Types of flow & Reynolds number

Water from a pipe is coming at a rate of 100 litres per minute. If the radius of the pipe is 5 cm, the Reynolds number for the flow is of the order of: (density of water = 1000 kg/m3, coefficient of viscosity of water = 1mPas)

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Answer: (b)

$R_e = \dfrac{\rho v D}{\eta}$ with $v = \dfrac{Q}{\pi r^2}$ , $D = 2r$ gives $R_e = \dfrac{2\rho Q}{\pi \eta r}$ .
$Q = 100\text{ L/min} = \dfrac{0.1}{60} \approx 1.67\times10^{-3}\text{ m}^3/\text{s}$
$R_e = \dfrac{2\times1000\times1.67\times10^{-3}}{\pi\times10^{-3}\times0.05} \approx 2.1\times10^{4}$
So the answer is (b) of the order of $10^4$ .

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Q19JEE Main 2019 · 10 APR · SHIFT 2Medium

Types of Flow, Reynolds Number & Continuity · Equation of continuity

Water from a tap emerges vertically downwards with an initial speed of 1.0 ms–1. The cross- sectional area of the tap is 10–4 m2. Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross-sectional area of the stream, 0.15 m below the tap would be: (Take g = 10 ms–2)

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Answer: (c)

Speed after falling $h$ : $v^2 = u^2 + 2gh = 1 + 2\times10\times0.15 = 4 \Rightarrow v = 2\text{ m/s}$
Continuity: $A_1 u = A_2 v \Rightarrow A_2 = \dfrac{10^{-4}\times1}{2}$
$A_2 = 5\times10^{-5}\text{ m}^2$
So the answer is (c).

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Q20JEE Main 2019 · 11 JAN · SHIFT 1Medium

Types of Flow, Reynolds Number & Continuity · Equation of continuity

A liquid of density $\rho$ is coming out of a hose pipe of radius a with horizontal speed $v$ and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% looses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be:

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Answer: (a)

Momentum carried per second by the jet $= \rho A v^2$ .
50% passes through: no momentum change.
25% stops: change $= 0.25\rho A v^2$ ; 25% rebounds with same speed: change $= 0.25\times 2\rho A v^2 = 0.5\rho A v^2$ .
$F = 0.75\rho A v^2 \Rightarrow P = \dfrac{F}{A} = \dfrac{3}{4}\rho v^2$
So the answer is (a).

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Q21JEE Main 2015 · 10 APR · SHIFT 1Medium

Types of Flow, Reynolds Number & Continuity · Types of flow & Reynolds number

If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter $\frac{2}{\sqrt{\pi}}$ cm then the Reynolds number for the flow is (density of water = 103 kg/m3 and viscosity of water = 10 $-$ 3 Pa.s) close to:

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Answer: (a)

$Q = \dfrac{15\times10^{-3}}{5\times60} = 5\times10^{-5}\text{ m}^3/\text{s}$
$r = \dfrac{1}{\sqrt{\pi}}\text{ cm} \Rightarrow A = \pi r^2 = 1\text{ cm}^2 = 10^{-4}\text{ m}^2$ , so $v = \dfrac{Q}{A} = 0.5\text{ m/s}$
$R_e = \dfrac{\rho v D}{\eta} = \dfrac{10^3 \times 0.5 \times \frac{2}{\sqrt{\pi}}\times10^{-2}}{10^{-3}} \approx 5.6\times10^{3}$
So the answer is (a) about $5500$ .

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Q22JEE Main 2011 · AIEEEMedium

Types of Flow, Reynolds Number & Continuity · Equation of continuity

Water is flowing continuously from a tap having an internal diameter $8 \times 10^{-3}\text{ m.}$ The water velocity as it leaves the tap is $0.4\text{ ms}^{-1}$ . The diameter of the water stream at a distance $2 \times 10^{-1}\text{ m}$ below the tap is close to:

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Answer: (c)

Speed at $h = 0.2\text{ m}$ below: $v^2 = (0.4)^2 + 2\times10\times0.2 = 4.16 \Rightarrow v \approx 2.04\text{ m/s}$
Continuity: $d_1^2 v_1 = d_2^2 v_2 \Rightarrow d_2 = d_1\sqrt{\dfrac{v_1}{v_2}}$
$d_2 = 8\times10^{-3}\sqrt{\dfrac{0.4}{2.04}} \approx 8\times10^{-3}\times0.44 \approx 3.6\times10^{-3}\text{ m}$
So the answer is (c).

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Q23MHT CET 2026 · 18 APR · SHIFT 2Medium

Bernoulli's Equation

An incompressible fluid (ideal fluid) is flowing through non uniform cross-sectional tube PQ as shown in the figure from end P to end Q. If $K_P$ and $K_Q$ are the kinetic energy per unit volume of the fluid at end P and end Q respectively, then

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Answer: (d)

From the figure $A_P < A_Q$ ; continuity $A_P v_P = A_Q v_Q$ gives $v_P > v_Q$ .
KE per unit volume $K = \frac{1}{2}\rho v^2$ , so $K_P > K_Q$ .
So the answer is (d).

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Q24MHT CET 2026 · 16 APR · SHIFT 2Medium

Bernoulli's Equation

A horizontal pipe carries water in a streamline flow. At point along the pipe, where the cross- sectional area is $A_1$ , the velocity of water is $V_1$ and the pressure is $P_1$ . What is the pressure of water at another point where the cross-sectional area is $A_2$ ?

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Answer: (b)

Continuity: $V_2 = \dfrac{A_1 V_1}{A_2}$
Bernoulli (horizontal): $P_2 = P_1 + \frac{1}{2}\rho (V_1^2 - V_2^2) = P_1 + \frac{1}{2}\rho V_1^2\left(1 - \dfrac{A_1^2}{A_2^2}\right)$
$P_2 = P_1 + \dfrac{\rho V_1^2}{2A_2^2}(A_2^2 - A_1^2)$
So the answer is (b).

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Q25MHT CET 2026 · 16 APR · SHIFT 1Medium

Bernoulli's Equation

A closed pipe containing liquid showed a pressure $P_1$ by gauge. When the valve was opened, pressure was reduced to $P_2$ . The speed of water flowing out of the pipe is ( $\rho$ =density of water)

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Answer: (b)

Closed valve: $v = 0$ , pressure $P_1$ . Open valve: pressure $P_2$ , speed $v$ .
Bernoulli: $P_1 = P_2 + \frac{1}{2}\rho v^2$
$v = \left[\dfrac{2(P_1 - P_2)}{\rho}\right]^{1/2}$
So the answer is (b).

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Q26NEET 2026 · RE-NEETMedium

Bernoulli's Equation

Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between $P$ and $Q$ is $15\text{ Nm}^{-2}$ . The area of cross-section at $P$ and $Q$ are $40\text{ cm}^2$ and $20\text{ cm}^2$ , respectively. The rate of flow of water through the pipe, in $\text{cm}^3\text{ s}^{-1}$ , is:

[Take density of water $= 1000\text{ kg m}^{-3}$ ]

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Answer: (a)

Continuity: $v_Q = \dfrac{40}{20}v_P = 2v_P$
Bernoulli: $P_P - P_Q = \frac{1}{2}\rho(v_Q^2 - v_P^2) = \frac{3}{2}\rho v_P^2$
$15 = 1500\,v_P^2 \Rightarrow v_P = 0.1\text{ m/s} = 10\text{ cm/s}$
$Q = A_P v_P = 40\times10 = 400\text{ cm}^3/\text{s}$
So the answer is (a) 400.

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Q27JEE Main 2026 · 8 APR · SHIFT 2Medium

Bernoulli's Equation

A liquid of density $600\text{ kg/m}^3$ flowing steadily in a tube of varying cross-section. The cross-section at a point $A$ is $1.0\text{ cm}^2$ and that at $B$ is $20\text{ mm}^2$ . Both the points $A$ and $B$ are in same horizontal plane, the speed of the liquid at $A$ is $10\text{ cm/s}$ . The difference in pressures at $A$ and $B$ points is ______ Pa.

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Answer: (d)

$A_A = 1.0\text{ cm}^2 = 100\text{ mm}^2$ , $A_B = 20\text{ mm}^2$ , so $v_B = 5v_A = 50\text{ cm/s} = 0.5\text{ m/s}$
Bernoulli: $P_A - P_B = \frac{1}{2}\rho(v_B^2 - v_A^2) = \frac{1}{2}\times600\times(0.25 - 0.01)$
$P_A - P_B = 300 \times 0.24 = 72\text{ Pa}$
So the answer is (d) 72.

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Q28JEE Main 2026 · 21 JAN · SHIFT 1Medium

Bernoulli's Equation

Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at $A$ and $B$ are $6\text{ cm}^2$ and $3\text{ cm}^2$ respectively. The rate of flow will be $\underline{\hspace{3em}}\ \text{cm}^3/\text{s}$ . (take $g = 10\text{ m/s}^2$ )

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Answer: (d)

Pressure difference from the columns: $P_A - P_B = \rho g h$ .
Bernoulli: $\frac{1}{2}(v_B^2 - v_A^2) = gh = 10\times0.05 = 0.5 \Rightarrow v_B^2 - v_A^2 = 1$
Continuity: $v_B = \dfrac{6}{3}v_A = 2v_A \Rightarrow 3v_A^2 = 1 \Rightarrow v_A = \dfrac{1}{\sqrt{3}}\text{ m/s} = \dfrac{100}{\sqrt{3}}\text{ cm/s}$
$Q = A_A v_A = 6\times\dfrac{100}{\sqrt{3}} = 200\sqrt{3}\text{ cm}^3/\text{s}$
So the answer is (d).

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Q29MHT CET 2025 · 25 APR · SHIFT 2Medium

Bernoulli's Equation

A horizontal pipeline carries water in a streamline flow. At a point along the pipe, where the cross- sectional area is $10\text{ cm}^2$ , the velocity of water is $1\text{ m/s}$ and pressure is 2000 Pa. The pressure of water at another point where the cross-sectional area $5\text{ cm}^2$ is

[Given $\rightarrow$ density of water $= 1000\text{ kg/m}^3$ ]

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Answer: (c)

Continuity: $v_2 = \dfrac{10}{5}\times 1 = 2\text{ m/s}$
Bernoulli: $P_2 = P_1 - \frac{1}{2}\rho (v_2^2 - v_1^2) = 2000 - \frac{1}{2}\times1000\times(4 - 1)$
$P_2 = 2000 - 1500 = 500\text{ Pa}$
So the answer is (c) 500 Pa.

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Q30JEE Main 2025 · 23 JAN · SHIFT 2Medium

Bernoulli's Equation

Water flows in a horizontal pipe whose one end is closed with a valve. The reading of the pressure gauge attached to the pipe is $P_1$ . The reading of the pressure gauge falls to $P_2$ when the valve is opened. The speed of water flowing in the pipe is proportional to

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Answer: (b)

Closed valve: $v = 0$ , gauge reads $P_1$ . Open valve: gauge reads $P_2$ with speed $v$ .
Bernoulli: $P_1 - P_2 = \frac{1}{2}\rho v^2 \Rightarrow v = \sqrt{\dfrac{2(P_1 - P_2)}{\rho}}$
So $v \propto \sqrt{P_1 - P_2}$ , answer (b).

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Q31JEE Main 2025 · 22 JAN · SHIFT 2Difficult

Bernoulli's Equation

A tube of length $L$ is shown in the figure. The radius of cross section at the point $(1)$ is 2 cm and at the point (2) is 1 cm, respectively. If the velocity of water entering at point (1) is $2\text{ m/s}$ , then velocity of water leaving the point (2) will be

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Answer: (d)

Continuity: $\pi r_1^2 v_1 = \pi r_2^2 v_2 \Rightarrow v_2 = \left(\frac{r_1}{r_2}\right)^2 v_1$
$v_2 = \left(\frac{2}{1}\right)^2 (2) = 8\text{ m/s}$
So the answer is (d) 8 m/s.

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Q32JEE Main 2024 · 8 APR · SHIFT 1Easy

Bernoulli's Equation

Correct Bernoulli's equation is (symbols have their usual meaning):

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Answer: (d)

Bernoulli: pressure energy + potential energy + kinetic energy per unit volume is constant.
$P + \rho g h + \frac{1}{2}\rho v^2 =$ constant
So the answer is (d).

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Q33MHT CET 2024 · 10 MAY · SHIFT 1Medium

Bernoulli's Equation

A closed pipe containing a liquid showed a pressure $P_1$ by gauge. When the valve was opened, pressure was reduced to $P_2$ . The speed of water flowing out of the pipe is ( $\rho =$ density of water)

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Answer: (c)

Closed valve: $v = 0$ , pressure $P_1$ . Open valve: pressure $P_2$ , speed $v$ .
Bernoulli: $P_1 - P_2 = \frac{1}{2}\rho v^2$
$v = \left[\dfrac{2(P_1 - P_2)}{\rho}\right]^{1/2}$
So the answer is (c).

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Q34MHT CET 2024 · 9 MAY · SHIFT 1Medium

Bernoulli's Equation

Water is flowing in a conical tube as shown in figure. Velocity of water at area ' $A_2$ ' is $60\text{ cm/s}$ . The value of ' $A_1$ ' and ' $A_2$ ' is $10\text{ cm}^2$ and $5\text{ cm}^2$ respectively. The pressure difference at both the cross- section is

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Answer: (c)

Continuity: $v_1 = \dfrac{A_2 v_2}{A_1} = \dfrac{5\times0.6}{10} = 0.3\text{ m/s}$
Bernoulli (horizontal): $P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2) = \frac{1}{2}\times1000\times(0.36 - 0.09)$
$P_1 - P_2 = 500 \times 0.27 = 135\text{ N/m}^2$
So the answer is (c).

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Q35MHT CET 2024 · 2 MAY · SHIFT 1Medium

Bernoulli's Equation

A streamline flow of a liquid of density ' $\rho$ ' is passing through a horizontal pipe of crosssectional area $A_1$ and $A_2$ at two ends. If the pressure of liquid is ' P ' at a point where flow speed is ' $v$ ', then pressure at another point where the flow of speed becomes 3 v is

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Answer: (d)

Bernoulli (horizontal): $P + \frac{1}{2}\rho v^2 = P' + \frac{1}{2}\rho (3v)^2$
$P' = P - \frac{1}{2}\rho(9v^2 - v^2) = P - 4\rho v^2$
So the answer is (d).

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Q36NEET 2024Medium

Bernoulli's Equation

An ideal fluid is flowing in a non-uniform cross-sectional tube $XY$ (as shown in the figure) from end $X$ to end $Y$ . If $K_1$ and $K_2$ are the kinetic energy per unit volume of the fluid at $X$ and $Y$ respectively, then the correct option is:

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Answer: (c)

From the figure, end $X$ is narrow and end $Y$ is wide: $A_X < A_Y$ .
Continuity $A_X v_X = A_Y v_Y$ gives $v_X > v_Y$ .
KE per unit volume $= \frac{1}{2}\rho v^2$ , so $K_1 > K_2$ (height does not affect it).
So the answer is (c).

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Q37MHT CET 2024 · 2 MAY · SHIFT 1Difficult

Bernoulli's Equation

Glycerine of density $1.25 \times 10^{3}\text{ kg/m}^3$ is flowing in conical shaped horizontal pipe. Crosssectional area of the pipe at its both ends is $10\text{ cm}^2$ and $5\text{ cm}^2$ respectively. Pressure difference at both the ends is $3\text{ N/m}^2$ . Rate of flow of liquid in the pipe is

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Answer: (a)

Continuity: $v_2 = \frac{10}{5}v_1 = 2v_1$
Bernoulli: $\Delta P = \frac{1}{2}\rho(4v_1^2 - v_1^2) \Rightarrow 3 = \frac{1}{2}(1250)(3v_1^2)$
$v_1^2 = 1.6\times10^{-3} \Rightarrow v_1 = 0.04\text{ m/s}$
$Q = A_1v_1 = (10\times10^{-4})(0.04) = 4\times10^{-5}\text{ m}^3\text{/s}$
So the answer is (a) $4\times10^{-5}\text{ m}^3\text{/s}$ .

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Q38MHT CET 2023 · 9 MAY · SHIFT 1Easy

Bernoulli's Equation

Water is flowing through a horizontal pipe in stream line flow. At the narrowest part of the pipe

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Answer: (a)

Continuity: $Av$ = constant, so the speed is largest where $A$ is smallest.
Bernoulli (horizontal): $P + \frac{1}{2}\rho v^2$ = constant, so larger $v$ means smaller $P$ .
So the answer is (a) velocity maximum, pressure minimum.

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Q39JEE Main 2023 · 13 APR · SHIFT 1Medium

Bernoulli's Equation

The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross - section. Cross sectional areas at $A$ is $1.5\text{ cm}^2$ , and $B$ is $25\text{ mm}^2$ , if the speed of liquid at $B$ is $60\text{ cm/s}$ then $(P_A - P_B)$ is:

(Given $P_A$ and $P_B$ are liquid pressures at $A$ and $B$ points.

density $\rho = 1000\text{ kg m}^{-3}$

$A$ and $B$ are on the axis of tube

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Answer: (b)

$A_A = 1.5\text{ cm}^2 = 150\text{ mm}^2$ , $A_B = 25\text{ mm}^2$ .
Continuity: $v_A = \dfrac{25}{150}\times 0.6 = 0.1\text{ m/s}$
Bernoulli: $P_A - P_B = \frac{1}{2}\times1000\times(0.36 - 0.01) = 500\times0.35$
$P_A - P_B = 175\text{ Pa}$
So the answer is (b).

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Q40JEE Main 2023 · 12 APR · SHIFT 1Medium

Bernoulli's Equation

Glycerin of density $1.25 \times 10^{3}\text{ kg m}^{-3}$ is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends are $10\text{ cm}^2$ and $5\text{ cm}^2$ and pressure drop across its length is $3\text{ Nm}^{-2}$ . The rate of flow of glycerin through the pipe is $x \times 10^{-5}\text{ m}^3\text{ s}^{-1}$ . The value of $x$ is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

Continuity: $A_1v_1 = A_2v_2 \Rightarrow v_2 = \frac{10}{5}v_1 = 2v_1$
Bernoulli (horizontal): $\Delta P = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2}\rho(3v_1^2)$
$3 = \frac{1}{2}(1250)(3v_1^2) \Rightarrow v_1^2 = 1.6\times10^{-3} \Rightarrow v_1 = 0.04\text{ m/s}$
$Q = A_1v_1 = (10\times10^{-4})(0.04) = 4\times10^{-5}\text{ m}^3\text{/s}$
So $x = 4$ .

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Q41JEE Main 2023 · 10 APR · SHIFT 2Difficult

Bernoulli's Equation

Figure below shows a liquid being pushed out of the tube by a piston having area of cross section $2.0\text{ cm}^2$ . The area of cross section at the outlet is $10\text{ mm}^2$ . If the piston is pushed at a speed of $4\text{ cm s}^{-1}$ , the speed of outgoing fluid is __________ $\text{cms}^{-1}$

Numerical answer type. Work it out, then open the answer.

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Answer: 80

Continuity: $A_1v_1 = A_2v_2$ , with $A_1 = 2.0\text{ cm}^2 = 200\text{ mm}^2$
$v_2 = \frac{200}{10}\times 4 = 80\text{ cm/s}$
So the answer is $80$ cm s $^{-1}$ .

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Q42JEE Main 2022 · 28 JUN · SHIFT 2Medium

Bernoulli's Equation

A liquid of density 750 kgm ${}^{-3}$ flows smoothly through a horizontal pipe that tapers in cross- sectional area from A1 = 1.2 $\times$ 10 $-$ 2 m2 to A2 = $\frac{A_1}{2}$ . The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is ___________ $\times$ 10 $-$ 3 m3s ${}^{-1}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 24

Continuity: $A_2 = \frac{A_1}{2} \Rightarrow v_2 = 2v_1$
Bernoulli: $\Delta P = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{3}{2}\rho v_1^2$
$4500 = \frac{3}{2}(750)v_1^2 = 1125\,v_1^2 \Rightarrow v_1 = 2\text{ m/s}$
$Q = A_1v_1 = (1.2\times10^{-2})(2) = 24\times10^{-3}\text{ m}^3\text{/s}$
So the answer is $24$ .

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Q43JEE Main 2022 · 26 JUN · SHIFT 1Medium

Bernoulli's Equation

An ideal fluid of density 800 kgm ${}^{-3}$ , flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to $\frac{a}{2}$ . The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is $\frac{\sqrt{x}}{6}$ ms ${}^{-1}$ for x = ___________. (Given g = 10 ms ${}^{-2}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 363

Continuity: area $a \to \frac{a}{2}$ , so $v_2 = 2v_1$ . The wide end is $h = 1$ m higher.
Bernoulli: $P_1 + \frac{1}{2}\rho v_1^2 + \rho g h = P_2 + \frac{1}{2}\rho v_2^2$
$(P_1 - P_2) + \rho g h = \frac{3}{2}\rho v_1^2 \Rightarrow 4100 + 800(10)(1) = \frac{3}{2}(800)v_1^2$
$12100 = 1200\,v_1^2 \Rightarrow v_1 = \frac{11}{\sqrt{12}} = \frac{11\sqrt{3}}{6} = \frac{\sqrt{363}}{6}\text{ m/s}$
So $x = 363$ .

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Q44JEE Main 2022 · 29 JUL · SHIFT 2Difficult

Bernoulli's Equation

A tube of length $50\text{ cm}$ is filled completely with an incompressible liquid of mass $250\text{ g}$ and closed at both ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity $x\sqrt{F}\text{ rad s}^{-1}$ . If $F$ be the force exerted by the liquid at the other end then the value of $x$ will be __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

The liquid moves in a circle; the end wall supplies the total centripetal force, acting on the centre of mass at $\frac{L}{2}$ .
$F = m\omega^2\frac{L}{2} = (0.25)\omega^2(0.25) = \frac{\omega^2}{16}$
$\omega = 4\sqrt{F}\text{ rad/s}$
So $x = 4$ .

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Q45JEE Main 2020 · 6 SEP · SHIFT 2Medium

Bernoulli's Equation

A fluid is flowing through a horizontal pipe of varying cross-section, with

speed v ms–1 at a point where the pressure is P pascal. At another point where pressure is $\frac{P}{2}$ Pascal its speed is V ms–1. If the density of the fluid is $\rho$ kg m–3 and the flow is streamline, then V is equal to:

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Answer: (b)

Bernoulli (horizontal): $P + \frac{1}{2}\rho v^2 = \dfrac{P}{2} + \frac{1}{2}\rho V^2$
$\dfrac{P}{2} = \frac{1}{2}\rho(V^2 - v^2) \Rightarrow V^2 = \dfrac{P}{\rho} + v^2$
So the answer is (b) $\sqrt{\dfrac{P}{\rho} + v^2}$ .

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Q46JEE Main 2020 · 9 JAN · SHIFT 1Medium

Bernoulli's Equation

Water flows in a horizontal tube (see figure).

The pressure of water changes by 700 Nm–2

between A and B where the area of cross section

are 40 cm2 and 20 cm2, respectively. Find the

rate of flow of water through the tube.

(density of water = 1000 kgm–3)

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Answer: (d)

Continuity: $v_B = \dfrac{40}{20}v_A = 2v_A$
Bernoulli: $P_A - P_B = \frac{1}{2}\rho(4v_A^2 - v_A^2) = 1500\,v_A^2 = 700$
$v_A^2 = 0.467 \Rightarrow v_A \approx 0.683\text{ m/s} = 68.3\text{ cm/s}$
$Q = A_A v_A = 40\times68.3 \approx 2732\text{ cm}^3/\text{s}$
So the answer is (d) $2720\text{ cm}^3/\text{s}$ .

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Q47NEET 2013Medium

Bernoulli's Equation

A fluid is in streamline flow across a horizontal pipe of variable area of cross section. For this which of the following statements is correct?

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Answer: (a)

Continuity: speed is maximum where the area is minimum (narrowest part).
Bernoulli (horizontal): pressure is maximum where speed is minimum (widest part).
So the answer is (a).

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Q48MHT CET 2025 · 23 APR · SHIFT 1Medium

Applications of Bernoulli's Theorem

A wind with speed $50\text{ m/s}$ blows parallel to the roof of a house. The area of the roof is $300\text{ m}^2$ . Assume that the pressure inside the house is atmospheric pressure. Density of air is $1.2\text{ kg/m}^3$ . The magnitude of the force exerted by the wind on the roof will be

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Answer: (c)

Air inside is at rest, air above moves at $v$ : $\Delta P = \frac{1}{2}\rho v^2$
$\Delta P = \frac{1}{2}(1.2)(50)^2 = 1500\text{ Pa}$
$F = \Delta P \cdot A = 1500 \times 300 = 4.5\times10^{5}\text{ N}$
So the answer is (c) $4.5\times10^{5}$ N.

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Q49JEE Main 2024 · 1 FEB · SHIFT 1Medium

Applications of Bernoulli's Theorem

A plane is in level flight at constant speed and each of its two wings has an area of $40\text{ m}^2$ . If the speed of the air is $180\text{ km/h}$ over the lower wing surface and $252\text{ km/h}$ over the upper wing surface, the mass of the plane is ___________ kg.

(Take air density to be $1\text{ kg m}^{-3}$ and $g = 10\text{ ms}^{-2}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 9600

$180\text{ km/h} = 50\text{ m/s}$ , $252\text{ km/h} = 70\text{ m/s}$
$\Delta P = \frac{1}{2}\rho(v_u^2 - v_l^2) = \frac{1}{2}(1)(4900 - 2500) = 1200\text{ Pa}$
Lift on both wings: $F = \Delta P \times 2A = 1200 \times 80 = 96000\text{ N}$
$mg = F \Rightarrow m = \frac{96000}{10} = 9600\text{ kg}$

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Q50JEE Main 2024 · 29 JAN · SHIFT 1Medium

Applications of Bernoulli's Theorem

In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are $70\text{ ms}^{-1}$ and $65\text{ ms}^{-1}$ respectively. If the wing area is $2\text{ m}^2$ , the lift of the wing is _________ $N$ .

(Given density of air $= 1.2\text{ kg m}^{-3}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 810

$\Delta P = \frac{1}{2}\rho(v_u^2 - v_l^2) = \frac{1}{2}(1.2)(70^2 - 65^2) = 0.6 \times 675 = 405\text{ Pa}$
Lift $= \Delta P \times A = 405 \times 2 = 810\text{ N}$

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Q51MHT CET 2023 · 13 MAY · SHIFT 1Easy

Applications of Bernoulli's Theorem

Venturimeter is used to

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Answer: (c)

A venturimeter measures the pressure difference between the wide and narrow (throat) sections.
Using Bernoulli's equation and continuity, this pressure difference gives the flow speed and hence the volume flow rate.
So the answer is (c) measure rate of flow of liquids.

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Q52NEET 2023Easy

Applications of Bernoulli's Theorem

The venturi-meter works on:

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Answer: (a)

A venturimeter relates the pressure drop at the throat to the flow speed using $P + \frac{1}{2}\rho v^2 = \text{constant}$ .
So the answer is (a) Bernoulli's principle.

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Q53JEE Main 2023 · 29 JAN · SHIFT 2Medium

Applications of Bernoulli's Theorem

A fully loaded boeing aircraft has a mass of $5.4 \times 10^{5}$ kg. Its total wing area is 500 m $^{2}$ . It is in level flight with a speed of 1080 km/h. If the density of air $\rho$ is 1.2 kg m $^{-3}$ , the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be. ( $g = 10\text{ m/s}^2$ )

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Answer: (d)

Level flight: lift $= mg \Rightarrow \Delta P = \frac{mg}{A} = \frac{5.4\times10^{6}}{500} = 1.08\times10^{4}\text{ Pa}$
Bernoulli: $\Delta P = \frac{1}{2}\rho(v_u^2 - v_l^2) \approx \rho v\,\Delta v$ , with $v = 1080\text{ km/h} = 300\text{ m/s}$
$\Delta v = \frac{1.08\times10^{4}}{1.2 \times 300} = 30\text{ m/s}$
$\frac{\Delta v}{v} = \frac{30}{300} = 0.1 = 10\%$
So the answer is (d) 10.

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Q54MHT CET 2019 · 2 MAY · SHIFT 2Easy

Applications of Bernoulli's Theorem

Two light balls are suspended as shown in figure. When a stream of air passes through the space between them, the distance between the balls will

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Answer: (d)

Air blown between the balls moves fast there, so by Bernoulli's principle the pressure between them drops.
The still air outside is at higher pressure and pushes the balls inward.
So the answer is (d) decrease.

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Q55NEET 2015 · AIPMT · CANCELLED PAPERMedium

Applications of Bernoulli's Theorem

A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m2. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be $(\rho_{air} = 1.2\, kg/m^3)$

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Answer: (a)

Pressure above the roof is lower: $\Delta P = \frac{1}{2}\rho v^2 = \frac{1}{2}(1.2)(40)^2 = 960\text{ Pa}$
$F = \Delta P \cdot A = 960 \times 250 = 2.4\times10^{5}\text{ N}$
Higher pressure is inside, so the force acts upwards.
So the answer is (a) $2.4\times10^{5}$ N, upwards.

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Q56JEE Main 2014 · 19 APR · SHIFT 1Medium

Applications of Bernoulli's Theorem

In the diagram shown, if the difference in the two tubes of the manometer is 5 cm, the cross section of the tube at A and B is 6 mm2 and 10 mm2 respectively. The rate at which water flows through the tube ( g = 10 ms ${}^{-2}$ )

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Answer: (a)

Manometer: $\Delta P = \rho g h = 1000(10)(0.05) = 500\text{ Pa}$
Bernoulli: $\frac{1}{2}\rho(v_A^2 - v_B^2) = \rho g h \Rightarrow v_A^2 - v_B^2 = 2gh = 1$
Continuity: $6v_A = 10v_B \Rightarrow v_B = 0.6v_A$ , so $0.64v_A^2 = 1 \Rightarrow v_A = 1.25\text{ m/s}$
$Q = A_Av_A = (6\times10^{-6})(1.25) = 7.5\times10^{-6}\text{ m}^3\text{/s} = 7.5\text{ cc/s}$
So the answer is (a) 7.5 cc/s.

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Q57JEE Main 2013 · 22 APR · SHIFT 1Medium

Applications of Bernoulli's Theorem

Air of density 1.2 kg m $^{-3}$ is blowing across the horizontal wings of an aeroplane in such a way that its speeds above and below the wings are 150 ms $^{-1}$ and 100 ms $^{-1}$ , respectively. The pressure difference between the upper and lower sides of the wings, is:

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Answer: (c)

Bernoulli: $\Delta P = \frac{1}{2}\rho(v_1^2 - v_2^2)$
$\Delta P = \frac{1}{2}(1.2)(150^2 - 100^2) = 0.6 \times 12500 = 7500\text{ N m}^{-2}$
So the answer is (c) 7500 N m $^{-2}$ .

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Q58MHT CET 2026 · 12 MAY · SHIFT 2Medium

Velocity of Efflux (Torricelli)

A water tank of height $\text{'}h\text{'}$ has a hole at the bottom. It is completely filled with water. The total pressure at the bottom is $4H$ and atmospheric pressure is $H$ . The velocity of water flowing out of the hole is ( $\rho$ = density of water)

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Answer: (a)

Gauge pressure at the bottom: $\rho g h = 4H - H = 3H$
Torricelli: $\frac{1}{2}\rho v^2 = \rho g h = 3H$
$v = \sqrt{\frac{6H}{\rho}}$
So the answer is (a) $\left(\frac{6H}{\rho}\right)^{1/2}$ .

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Q59MHT CET 2026 · 13 APR · SHIFT 2Medium

Velocity of Efflux (Torricelli)

The surface of water in a water tank of cross-section area $750$ cm $^{2}$ on the top of a house is $h$ m above the tap level. The speed of water coming out through the tap of cross-section area $500$ mm $^{2}$ is $30$ cm/s. At that instant, $dh/dt$ is $y \times 10^{-3}$ m/s. The value of $y$ will be

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Answer: (a)

Continuity: $A\left|\frac{dh}{dt}\right| = a v$
$\left|\frac{dh}{dt}\right| = \frac{(500\times10^{-6})(0.30)}{750\times10^{-4}} = \frac{1.5\times10^{-4}}{0.075} = 2\times10^{-3}\text{ m/s}$
So the answer is (a) $y = 2$ .

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Q60JEE Main 2026 · 6 APR · SHIFT 2Medium

Velocity of Efflux (Torricelli)

A cylindrical vessel of 40 cm radius is completely filled with water and its capacity is $528\text{ dm}^3$ (dm: decimeter) The vessel is placed on a solid block of exactly same height as vessel. If a small hole is made at 70 cm below the top of water level, then horizontal range of water falling on the ground in the beginning is ______ cm.

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Answer: (b)

Height of vessel: $H = \frac{V}{\pi r^2} = \frac{0.528}{\frac{22}{7}(0.4)^2} = 1.05\text{ m} = 105\text{ cm}$
Block has the same height, so the water top is 210 cm above the ground; the hole is at $y = 210 - 70 = 140$ cm above ground, depth $h = 70$ cm.
Range: $R = 2\sqrt{h\,y} = 2\sqrt{70 \times 140} = 2 \times 70\sqrt{2} = 140\sqrt{2}\text{ cm}$
So the answer is (b) $140\sqrt{2}$ .

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Q61MHT CET 2026 · 20 APR · SHIFT 2Difficult

Velocity of Efflux (Torricelli)

On opposite sides of a wide vertical vessel filled with water (density $\rho$ ), two identical holes are drilled, each having cross-section area $A$ . The height difference between holes is $x$ . The resultant force of reaction of water flowing out of vessel is ( $g$ = acceleration due to gravity)

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Answer: (b)

Reaction force of a jet: $F = \rho A v^2 = \rho A(2gh) = 2\rho A g h$
The jets leave in opposite directions, so the reactions oppose each other.
Net force: $F_2 - F_1 = 2\rho A g(h_2 - h_1) = 2\rho A g x$
So the answer is (b) $2\rho A g x$ .

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Q62MHT CET 2025 · 5 MAY · SHIFT 2Difficult

Velocity of Efflux (Torricelli)

A vessel completely filled with water has two holes ' P ' and ' Q ' at depths ' 2 h ' and ' 8 h ' from the top respectively. Hole ' P ' is square of side ' $a$ ' and hole ' $Q$ ' is a circle of radius ' $r$ '. The water flowing out per second from both the holes is same, then side ' $a$ ' of hole ' $P$ ' is

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Answer: (b)

Volume flow rate $Q = (\text{area}) \times v$ with efflux speed $v = \sqrt{2gd}$ at depth $d$ .
Equal flow: $a^2\sqrt{2g(2h)} = \pi r^2\sqrt{2g(8h)}$
$a^2 = \pi r^2\sqrt{\frac{8h}{2h}} = 2\pi r^2$
So the answer is (b) $a = r\sqrt{2\pi}$ .

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Q63MHT CET 2025 · 22 APR · SHIFT 1Difficult

Velocity of Efflux (Torricelli)

There are two identical small holes on the opposite side of a tank containing full of a liquid. The tank is open at the top. The difference in height between the two holes is ' $h$ '. As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to

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Answer: (d)

Each jet pushes the tank back with thrust $F = \rho a v^2$ , and $v^2 = 2gd$ at depth $d$ .
The two jets point in opposite directions, so $F_{net} = \rho a(v_2^2 - v_1^2) = \rho a \cdot 2g(d_2 - d_1)$
$d_2 - d_1 = h \Rightarrow F_{net} = 2\rho a g h$
So the answer is (d) $F_{net} \propto h$ .

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Q64NEET 2025Difficult

Velocity of Efflux (Torricelli)

Consider a water tank shown in the figure. It has one wall at $x = L$ and can be taken to be very wide in the $z$ direction. When filled with a liquid of surface tension $S$ and density $\rho$ , the liquid surface makes angle $\theta_0\ (\theta_0 \ll 1)$ with the $x$ -axis at $x = L$ . If $y(x)$ is the height of the surface then the equation for $y(x)$ is:

(take $\theta(x) = \sin\theta(x) = \tan\theta(x) = \frac{dy}{dx}, g$ is the acceleration due to gravity)

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Answer: (d)

For small slope, curvature of the surface is $\frac{1}{R} = \frac{d^2y}{dx^2}$ , so the pressure jump across it is $\frac{S}{R} = S\frac{d^2y}{dx^2}$ .
Liquid just below the surface at height $y$ (above the flat level) has pressure $p_0 - \rho g y$ .
Balance: $p_0 - (p_0 - \rho g y) = S\frac{d^2y}{dx^2}$
So the answer is (d) $\frac{d^2y}{dx^2} = \frac{\rho g}{S}y$ .

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Q65JEE Main 2025 · 3 APR · SHIFT 1Difficult

Velocity of Efflux (Torricelli)

Consider a completely full cylindrical water tank of height 1.6 m and of cross-sectional area $0.5\text{ m}^2$ . It has a small hole in its side at a height 90 cm from the bottom. Assume, the cross-sectional area of the hole to be negligibly small as compared to that of the water tank. If a load 50 kg is applied at the top surface of the water in the tank then the velocity of the water coming out at the instant when the hole is opened is:

$(g = 10\text{ m/s}^2)$

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Answer: (d)

Extra pressure from load: $P = \frac{mg}{A} = \frac{50 \times 10}{0.5} = 1000\text{ Pa}$
Depth of hole: $h = 1.6 - 0.9 = 0.7\text{ m}$
Bernoulli: $\frac{1}{2}\rho v^2 = \rho g h + P \Rightarrow v^2 = 2\left(10 \times 0.7 + \frac{1000}{1000}\right) = 16$
So the answer is (d) $v = 4\text{ m/s}$ .

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Q66MHT CET 2024 · 16 MAY · SHIFT 1Easy

Velocity of Efflux (Torricelli)

A cylinder contains water upto a height ' $H$ '. It has three orifices $O_1, O_2, O_3$ as shown in the figure. Let $V_1, V_2, V_3$ be the speed of efflux of water from the three orifices. Then

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Answer: (b)

Torricelli: $v = \sqrt{2gh}$ , where $h$ is the depth of the orifice below the free surface.
$O_1$ is shallowest and $O_3$ deepest, so $h_1 < h_2 < h_3$ .
So the answer is (b) $V_1 < V_2 < V_3$ .

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Q67MHT CET 2024 · 10 MAY · SHIFT 1Medium

Velocity of Efflux (Torricelli)

A completely filled water tank of height ' $h$ ' has a hole at the bottom. The total pressure of the bottom is 4 H and atmospheric pressure is H. The velocity of water flowing out of the hole is ( $\rho =$ density of water)

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Answer: (c)

Gauge pressure at the bottom: $\rho g h = 4H - H = 3H$
$\frac{1}{2}\rho v^2 = 3H \Rightarrow v = \sqrt{\frac{6H}{\rho}}$
So the answer is (c) $\sqrt{\frac{6H}{\rho}}$ .

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Q68MHT CET 2024 · 4 MAY · SHIFT 1Medium

Velocity of Efflux (Torricelli)

The surface of water in a water tank of cross section area $750\text{ cm}^2$ on the top of a house is ' $h$ ' $\text{m}$ above the tap level. The speed of water coming out through the tap of cross section area $500\text{ mm}^2$ is $30\text{ cm/s}$ . At that instant $\frac{dh}{dt}$ is $x \times 10^{-3}\text{ m/s}$ . The value of ' $x$ ' will be

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Answer: (a)

Continuity: $A\left|\frac{dh}{dt}\right| = a v$
$\left|\frac{dh}{dt}\right| = \frac{(500\times10^{-6})(0.30)}{750\times10^{-4}} = 2\times10^{-3}\text{ m/s}$
So the answer is (a) $x = 2$ .

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Q69MHT CET 2024 · 3 MAY · SHIFT 1Medium

Velocity of Efflux (Torricelli)

A gardening pipe having an internal radius ' $R$ ' is connected to a water sprinkler having ' $n$ ' holes each of radius ' $r$ '. The water in the pipe has a speed ' $v$ '. The speed of water leaving the sprinkler is

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Answer: (b)

Continuity: flow in the pipe = total flow out of $n$ holes
$\pi R^2 v = n\pi r^2 v' \Rightarrow v' = \frac{R^2 v}{n r^2}$
So the answer is (b) $\frac{R^2 v}{n r^2}$ .

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Q70JEE Main 2024 · 27 JAN · SHIFT 2Medium

Velocity of Efflux (Torricelli)

The reading of pressure metre attached with a closed pipe is $4.5 \times 10^{4}\text{ N/m}^2$ . On opening the valve, water starts flowing and the reading of pressure metre falls to $2.0 \times 10^{4}\text{ N/m}^2$ . The velocity of water is found to be $\sqrt{V}\text{ m/s}$ . The value of $V$ is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 50

Fall in static pressure becomes kinetic energy: $\Delta P = \frac{1}{2}\rho v^2$
$4.5\times10^{4} - 2.0\times10^{4} = \frac{1}{2}(1000)v^2 \Rightarrow v^2 = 50$
So $V = 50$ .

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Q71MHT CET 2023 · 13 MAY · SHIFT 2Medium

Velocity of Efflux (Torricelli)

There is hole of area $A$ at the bottom of a cylindrical vessel. Water is filled to a height $h$ and water flows out in $t$ second. If water is filled to a height $4h$ , it will flow out in time (in second)

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Answer: (c)

Time to empty: $t = \frac{A_{vessel}}{A}\sqrt{\frac{2h}{g}} \Rightarrow t \propto \sqrt{h}$
$\frac{t'}{t} = \sqrt{\frac{4h}{h}} = 2$
So the answer is (c) $2t$ .

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Q72JEE Main 2023 · 1 FEB · SHIFT 2Medium

Velocity of Efflux (Torricelli)

The surface of water in a water tank of cross section area $750\text{ cm}^2$ on the top of a house is $h\text{ m}$ above the tap level. The speed of water coming out through the tap of cross section area $500\text{ mm}^2$ is $30\text{ cm/s}$ . At that instant, $\frac{dh}{dt}$ is $x \times 10^{-3}\text{ m/s}$ . The value of $x$ will be ____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

Continuity: $A\left|\frac{dh}{dt}\right| = a v$
$\left|\frac{dh}{dt}\right| = \frac{(500\times10^{-6})(0.30)}{750\times10^{-4}} = 2\times10^{-3}\text{ m/s}$
So $x = 2$ .

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Q73JEE Main 2022 · 28 JUL · SHIFT 2Medium

Velocity of Efflux (Torricelli)

Consider a cylindrical tank of radius $1\text{ m}$ is filled with water. The top surface of water is at $15\text{ m}$ from the bottom of the cylinder. There is a hole on the wall of cylinder at a height of $5\text{ m}$ from the bottom. A force of $5 \times 10^{5}\text{ N}$ is applied on the top surface of water using a piston. The speed of efflux from the hole will be: (given atmospheric pressure $P_A = 1.01 \times 10^{5}\text{ Pa}$ , density of water $\rho_W = 1000\text{ kg/m}^3$ and gravitational acceleration $g = 10\text{ m/s}^2$ )

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Answer: (c)

Extra pressure from the piston: $\frac{F}{\pi r^2} = \frac{5\times10^{5}}{\pi(1)^2} \approx 1.59\times10^{5}\text{ Pa}$
Full Bernoulli (top to hole, depth 10 m): $\frac{1}{2}\rho v^2 = \frac{F}{A} + \rho g(10) = 1.59\times10^{5} + 1.0\times10^{5} \Rightarrow v \approx 22.8\text{ m/s}$ (not among the options)
The official answer uses only the piston pressure: $\frac{1}{2}\rho v^2 = 1.59\times10^{5} \Rightarrow v = \sqrt{318} \approx 17.8\text{ m/s}$
So the answer is (c) 17.8 m/s.

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Q74JEE Main 2022 · 27 JUN · SHIFT 1Medium

Velocity of Efflux (Torricelli)

The area of cross-section of a large tank is 0.5 m2. It has a narrow opening near the bottom having area of cross-section 1 cm2. A load of 25 kg is applied on the water at the top in the tank. Neglecting the speed of water in the tank, the velocity of the water, coming out of the opening at the time when the height of water level in the tank is 40 cm above the bottom, will be ___________ cms ${}^{-1}$ . [Take g = 10 ms ${}^{-2}$ ]

Numerical answer type. Work it out, then open the answer.

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Answer: 300

Extra pressure from the load: $\frac{mg}{A} = \frac{25 \times 10}{0.5} = 500\text{ Pa}$
Bernoulli: $\frac{1}{2}\rho v^2 = \frac{mg}{A} + \rho g h = 500 + 1000(10)(0.4) = 4500$
$v^2 = 9 \Rightarrow v = 3\text{ m/s} = 300\text{ cm/s}$

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Q75JEE Main 2021 · 27 JUL · SHIFT 2Medium

Velocity of Efflux (Torricelli)

The water is filled upto height of 12 m in a tank having vertical sidewalls. A hole is made in one of the walls at a depth 'h' below the water level. The value of 'h' for which the emerging steam of water strikes the ground at the maximum range is ________ m.

Numerical answer type. Work it out, then open the answer.

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Answer: 6

Hole at depth $h$ , height above ground $H - h$ : range $R = 2\sqrt{h(H - h)}$
$R$ is maximum when $h = H - h \Rightarrow h = \frac{H}{2}$
$h = \frac{12}{2} = 6\text{ m}$

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Q76JEE Main 2021 · 18 MAR · SHIFT 2Medium

Velocity of Efflux (Torricelli)

Consider a water tank as shown in the figure. It's cross-sectional area is 0.4 m2. The tank has an opening B near the bottom whose cross-section area is 1 cm2. A load of 24 kg is applied on the water at the top when the height of the water level is 40 cm above the bottom, the velocity of water coming out the opening B is v ms-1.

The value of v, to the nearest integer, is ___________. [Take value of g to be 10 ms-2]

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Extra pressure from the load: $\frac{mg}{A} = \frac{24 \times 10}{0.4} = 600\text{ Pa}$
Bernoulli: $\frac{1}{2}\rho v^2 = 600 + 1000(10)(0.4) = 4600$
$v^2 = 9.2 \Rightarrow v \approx 3.03\text{ m/s}$
So $v \approx 3$ .

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Q77JEE Main 2021 · 27 JUL · SHIFT 1Difficult

Velocity of Efflux (Torricelli)

A light cylindrical vessel is kept on a horizontal surface. Area of base is A. A hole of cross-sectional area 'a' is made just at its bottom side. The minimum coefficient of friction necessary to prevent sliding the vessel due to the impact force of the emerging liquid is (a < < A):

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Answer: (c)

Efflux speed $v = \sqrt{2gh}$ , so the jet's reaction force is $F = \rho a v^2 = 2\rho a g h$ .
Vessel is light, so normal force = weight of liquid: $N = \rho A h g$ .
No sliding: $\mu N \ge F \Rightarrow \mu \ge \frac{2\rho a g h}{\rho A h g} = \frac{2a}{A}$
So the answer is (c) $\frac{2a}{A}$ .

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Q78MHT CET 2020 · 16 OCT · SHIFT 2Medium

Velocity of Efflux (Torricelli)

A large open tank containing water has two holes to its wall. A square hole of side $a$ is made at a depth $y$ and a circular hole of radius $r$ is made at a depth $16y$ from the surface of water. If equal amount of water comes out through both the holes per second, then the relation between $r$ and $a$ will be

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Answer: (a)

Equal flow: $a^2\sqrt{2gy} = \pi r^2\sqrt{2g(16y)}$
$a^2 = 4\pi r^2 \Rightarrow r = \frac{a}{2\sqrt{\pi}}$
So the answer is (a) $r = \frac{a}{2\sqrt{\pi}}$ .

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Q79NEET 2019Easy

Velocity of Efflux (Torricelli)

A small hole of area of cross-section 2 mm2 present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s2, the rate of flow of water through the open hole would be nearly:

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Answer: (d)

Torricelli: $v = \sqrt{2gh} = \sqrt{2(10)(2)} = \sqrt{40} \approx 6.32\text{ m/s}$
$Q = av = (2\times10^{-6})(6.32) \approx 12.6\times10^{-6}\text{ m}^3\text{/s}$
So the answer is (d) $12.6\times10^{-6}$ m $^3$ /s.

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Q80JEE Main 2019 · 10 JAN · SHIFT 1Medium

Velocity of Efflux (Torricelli)

Water flows into a large tank with flat bottom at the rate of 10–4m3s–1. Water is also leaking out of a hole ofarea 1 cm2 at its bottom. If the height of the water in the tank remains steady, then this height is -

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Answer: (b)

Steady level: inflow = outflow, $Q = a\sqrt{2gh}$
$\sqrt{2gh} = \frac{10^{-4}}{10^{-4}} = 1\text{ m/s}$
$h = \frac{1}{2(9.8)} \approx 0.051\text{ m} = 5.1\text{ cm}$
So the answer is (b) 5.1 cm.

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Q81JEE Main 2019 · 9 JAN · SHIFT 2Medium

Velocity of Efflux (Torricelli)

The top of a water tank is open to air and its water level is mainted. It is giving out 0.74 m3 water per minute through a circular opening of 2 cm radius in its wall. The depth of the center of the opening from the level of water in the tank is close to:

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Answer: (b)

$Q = \frac{0.74}{60} = 1.23\times10^{-2}\text{ m}^3\text{/s}$ , $a = \pi(0.02)^2 = 1.26\times10^{-3}\text{ m}^2$
$v = \frac{Q}{a} \approx 9.8\text{ m/s}$
$h = \frac{v^2}{2g} = \frac{96}{2(10)} \approx 4.8\text{ m}$
So the answer is (b) 4.8 m.

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Q82JEE Main 2016 · 9 APR · SHIFT 1Difficult

Velocity of Efflux (Torricelli)

Consider a water jar of radius R that has water filled up to height H and is kept on astand of height h (see figure). Through a hole of radius r (r << R) at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is x. Then:

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Answer: (c)

Speed at the hole: $v_1 = \sqrt{2gH}$ ; speed at the ground: $v_2 = \sqrt{2g(H+h)}$ .
Continuity: $\pi r^2 v_1 = \pi x^2 v_2 \Rightarrow x^2 = r^2\sqrt{\frac{H}{H+h}}$
So the answer is (c) $x = r\left(\frac{H}{H+h}\right)^{1/4}$ .

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Q83NEET 2015 · AIPMTMedium

Velocity of Efflux (Torricelli)

The cylindrical tube of a spray pump has radius R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is V, the speed of the ejection of the liquid through the holes is

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Answer: (d)

Continuity: $\pi R^2 V = n\pi r^2 v$
$v = \frac{VR^2}{nr^2}$
So the answer is (d) $\frac{VR^2}{nr^2}$ .

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Q84JEE Main 2014 · 12 APR · SHIFT 1Medium

Velocity of Efflux (Torricelli)

A cylindrical vessel of cross-section A contains water to a height h. There is a hole in the bottom of radius 'a'. The time in which it will be emptied is:

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Answer: (b)

Continuity: $-A\frac{dh}{dt} = \pi a^2\sqrt{2gh}$
Integrate from $h$ to 0: $t = \frac{A}{\pi a^2}\cdot\frac{2\sqrt{h}}{\sqrt{2g}} = \frac{A}{\pi a^2}\sqrt{\frac{2h}{g}}$
$t = \frac{\sqrt{2}A}{\pi a^2}\sqrt{\frac{h}{g}}$
So the answer is (b).

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Q85JEE Main 2014 · 11 APR · SHIFT 1Medium

Velocity of Efflux (Torricelli)

A tank with a small hole at the bottom has been filled with water and kerosene (specific gravity 0.8). The height of water is 3 m and that of kerosene 2 m. When the hole is opened the velocity of fluid coming out from it is nearly: (take g = 10 ms ${}^{-2}$ and density of water = 103 kg m ${}^{-3}$ )

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Answer: (b)

Gauge pressure at the bottom: $\rho_k g h_k + \rho_w g h_w = 800(10)(2) + 1000(10)(3) = 46000\text{ Pa}$
Bernoulli at the hole: $\frac{1}{2}\rho_w v^2 = 46000 \Rightarrow v^2 = 92$
$v \approx 9.6\text{ m/s}$
So the answer is (b) 9.6 m s $^{-1}$ .

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Q86JEE Main 2002 · AIEEEEasy

Velocity of Efflux (Torricelli)

A cylinder of height $20\text{ m}$ is completely filled with water. The velocity of efflux of water (in $\text{ms}^{-1}$ ) through a small hole on the side wall of the cylinder near its bottom is

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Answer: (b)

Torricelli: $v = \sqrt{2gh} = \sqrt{2(9.8)(20)} = \sqrt{392} \approx 19.8 \approx 20\text{ m/s}$
So the answer is (b) 20.

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